In the circuit shown in the figure,the power | vedclass

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(D) The power dissipated in the $6\,\Omega$ resistor is given by $P = I^2 R_{6\Omega}$.Given $P = 6\,W$ and $R_{6\Omega} = 6\,\Omega$,we have $6 = I^2

Vedclass · Accepted Answer

$24$

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(D) The power dissipated in the $6\,\Omega$ resistor is given by $P = I^2 R_{6\Omega}$.Given $P = 6\,W$ and $R_{6\Omega} = 6\,\Omega$,we have $6 = I^2 	imes 6$,which implies $I^2 = 1$,so the current $I = 1\,A$.This current $I$ flows through the $6\,\Omega$ resistor,which is in series with the parallel combination of $R$ and $8\,\Omega$.The equivalent resistance of the parallel part is $R_p = \frac{8R}{8+R}$.The total resistance of the circuit is $R_{eq} = 6 + \frac{8R}{8+R} = \frac{6(8+R) + 8R}{8+R} = \frac{48 + 6R + 8R}{8+R} = \frac{48 + 14R}{8+R}$.Using Ohm's law,$V = I R_{eq}$,where $V = 12\,V$ and $I = 1\,A$:$12 = 1 	imes \frac{48 + 14R}{8+R}$$12(8 + R) = 48 + 14R$$96 + 12R = 48 + 14R$$96 - 48 = 14R - 12R$$48 = 2R$$R = 24\,\Omega$.

In the circuit shown in the figure,the power dissipated as heat in the $6\,\Omega$ resistor is $6\,W$. What is the value of resistance $R$ in the circuit? $\Omega$

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